Let $R$ be the region enclosed by the polar curve $r(\theta)=4+2\sin^2(3\theta)$ where $\dfrac{\pi}{4}\leq \theta\leq \pi$. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{\scriptsize\dfrac{\pi}{4}}^{\scriptsize\dfrac{3\pi}{4}}\left( 8+8\sin^2(3\theta)+2\sin^4(3\theta)\right)d\theta$ (Choice B) B $ \int_{\scriptsize\dfrac{\pi}{4}}^{\pi}\left( 2+\sin^2(3\theta)\right)d\theta$ (Choice C) C $ \int_{\scriptsize\dfrac{\pi}{4}}^{\scriptsize\dfrac{3\pi}{4}}\left( 2+\sin^2(3\theta)\right)d\theta$ (Choice D) D $ \int_{\scriptsize\dfrac{\pi}{4}}^{\pi}\left( 8+8\sin^2(3\theta)+2\sin^4(3\theta)\right)d\theta$
This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ Let's plug ${r(\theta)=4+2\sin^2(3\theta)}$, ${\alpha=\dfrac{\pi}{4}}$, and ${\beta=\pi}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize\dfrac{\pi}{4}}}^{{\pi}}\dfrac{1}{2}\left({4+2\sin^2(3\theta)}\right)^{2}d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{4}}^{\pi}\dfrac{1}{2}\left( 16+16\sin^2(3\theta)+4\sin^4(3\theta)\right)d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{4}}^{\pi}\left( 8+8\sin^2(3\theta)+2\sin^4(3\theta)\right)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize\dfrac{\pi}{4}}^{\pi}\left( 8+8\sin^2(3\theta)+2\sin^4(3\theta)\right)d\theta$